Chemistry 232

Chemical Kinetics

Chemical Kinetics

Chemical kinetics - speed or rate at which a reaction occurs

How are rates of reactions affected by• Reactant concentration?

• Temperature?

• Reactant states?

• Catalysts?

The Instantaneous Reaction Rate

Consider the following reaction

A + B CDefine the instantaneous rate of

consumption of reactant A, A

dt

AdA

Reaction Rates and Reaction Stoichiometry

Look at the reaction

O3(g) + NO(g) NO2(g) + O2(g)

dt

]Od[+ =

dt]NOd[

+ = dt

d[NO]- =

dtOd

- = rate 223

Another Example

2 NOCl (g) 2 NO + 1 Cl2 (g)

dtd[Cl

+ = dt

d[NO]21

= dtNOCld

21

- = rate 2 ]

WHY? 2 moles of NOCl disappear for every 1 mole Cl2 formed.

The General Case

a A + b B c C + d D

rate = -1 d[A] = -1 d[B] = +1 d[C] = +1 d[D] a dt b dt c dt d dt

Why do we define our rate in this way?

removes ambiguity in the measurement of reaction rates

Obtain a single rate for the entire equation, not just for the change in a single reactant or product.

Alternative Definition of the Rate

Rate of conversion related to the advancement of the reaction, .

dtd

V1

= reaction of rate

V = solution volume

An Example

Examine the following reaction

2 N2O5 (g) 4 NO2 (g) + O2 (g)

N2O5 NO2 O2

Initial nø 0 0

Change -2 +4 +

Final nø - 2 4

The N2O5 Decomposition

dt

Vd =

dt]d[O

dtVd

4 = dt

]d[NO

dt

Vd2- =

dtONd

2

2

52

dtd

V1

= reaction of rate

Note – constant volume system

The Rate Law

Relates rate of the reaction to the reactant concentrations and rate constant

For a general reaction a A + b B + c C d D + e E

zyx [C][B]k[A] dtd

V1

v Rate

Rate Laws (Cont’d)

The only way that we can determine the superscripts (x, y, and z) for a non-elementary chemical reaction is by experimentation.

Use the isolation method (see first year textbooks).

For a general reaction

x + y + z = reaction order

e.g. X = 1; Y = 1; Z = 0

2nd order reaction (x + y + z = 2)

X = 0; Y = 0; Z = 1 (1st order reaction)

X = 2; Y = 0; Z = 0 (2nd order)

Rate Laws for Multistep Processes

Chemical reactions generally proceed via a large number of elementary steps - the reaction mechanism

The slowest elementary step the rate determining step (rds).

An Example Reaction Mechanism

O atoms are intermediates in the above reaction sequence.• Intermediates – generally small, indeterminate

concentrations.

1

1

2

3 2

3 2

O O O

O O 2O

k

k

k

Integrated Rate Laws

The rate law gives us information about how the concentration of the reactant varies with time

How much reactant remains after specified period of time? Use the integrated rate laws.

First Order Reaction

A product Rate = v = - d[A]/dt = k[A]

How does the concentration of the reactant depend on time?

ktAA

o

ln

k has units of s-1

The Half-life of a First Order Reaction

For a first order reaction, the half-life t1/2

is calculated as follows.

k6930

t 21

./

Radioactive Decay

Radioactive Samples decay according to first order kinetics.

This is the half-life of samples containing e.g. 14C , 239Pu, 99Tc.

Example

011414 NC

Second Order Reaction

A products v = k[A]2

A + B products v = k[A][B]Case 1 is 2nd order in ACase 2 is 1st order in A and B and 2nd

order overall

The Dependence of Concentration on Time

For a second order process where v = k[A]2

ktA1

A1

o

Half-life for This Second Order Reaction.

[A] at t = t½ = ½ [A]0

021

21o0

Ak1

= t or

kt + A1

= /2A

1

][

][][

/

/

Other Second Order Reactions

Examine the Case 2 from above

A + B products v = k[A][B]

oo

o

o ABkt

AA

BB

ln

A Pseudo-first Order Reaction

Example hydration of methyl iodide

CH3I(aq) + H2O(l) CH3OH(aq) + H+(aq) + I-(aq)

Rate = k [CH3I] [H2O]

Since we carry out the reaction in aqueous solution

[H2O] >>>> [CH3I] [H2O] doesn’t change by a lot

Pseudo-first Order (cont’d)

Since the concentration of H2O is essentially constant

v = k [CH3I] [constant]

= k`[CH3I] where k` = k [H2O]

The reaction is pseudo first order since it appears to be first order, but it is actually a second order process.

Sequential First Order Reactions

Suppose we have two first order reactions occurring in sequence.

v k A

v k A

b a

c b

A B

B C

What is Our Intermediate?

B is an intermediate in the above reaction sequence.• Clearly B is formed in the first elementary

step of the reaction mechanism and consumed in the final step.

The Concentration Dependencies of the Species

The amounts of the reactants are related to the reaction rates as follows.

;a b

a b

d A d Ck A k B

dt dtd B

k A k Bdt

Sequential Reactions (3)

For a set of initial conditions [A]o 0, and [B]o

= [C]o =0 mol/L.

k

k

A A

Ak A k B

a

a

t

O

ta bO

e

de

dt

Sequential Reactions (4)

The concentration of the intermediate can be written

k kkB A

k ka bt ta

Ob a

e e

Sequential Reactions (5)

k kk kC 1 A

k k

b at ta b

Ob a

e e

o

o

Note A A B C

Hence C A A B

The Rate Determining Step of the Reaction

What happens if one of the steps in the reaction is much slower than the other reaction step?

k k

k

k klim lim 1 A

k k

1 A

b a

b b

a

t ta b

Ok kb a

t

O

e eC

e

Note – assuming ka <<< kb

The Rate Determining Step of the Reaction (2)

If kb <<< ka

k k

k

k klim lim 1 A

k k

1 A

b a

a a

b

t ta b

Ok kb a

t

O

e eC

e

Temperature Dependence of Reaction Rates

Reaction rates generally increase with increasing temperature.

Arrhenius Equation

RTEa

e Ak

A = pre-exponential factorEa = the activation energy

Reactions Approaching Equilibrium

Examine the concentration profiles for the following simple process.

A ⇌ B

Time

Co

nce

ntr

atio

n A

B

Approaching Equilibrium

Calculate the amounts of A and B at equilibrium.

k

vVAA eqA

oeq,

/

,

k

vVB eqB

eq

The Equilibrium Condition

At equilibrium, vA,eq = vB,eq.

eqBeqeqAeq vBkvAk ,/

,

KA

B

kk

eq

eq /

Elementary steps and the Molecularity

Kinetics of the elementary step depends only on the number of reactant molecules in that step!

• Molecularity the number of reactant molecules that participate in elementary steps

The Kinetics of Elementary Steps

For the elementary step

• unimolecular step For elementary steps involving more than one

reactant

• a bimolecular step

Akv productsA

B Akv productsBA

The Kinetics of Elementary Steps

For the elementary step

• unimolecular step For elementary steps involving more than one

reactant

• a bimolecular step

Akv productsA

B Akv productsBA

For the step

• a termolecular (three molecule) step.

• Termolecular (and higher) steps are not that common in reaction mechanisms.

BAkv productsBA2 2

The Steady-State Approximation

Examine the following simple reaction mechanism

Bkv 2p

Rate of product formation, vp, is proportional to the concentration of an intermediate.

1 2

1

k k

kA B P

⇌

Applying the Steady State Approximation (SSA)

Look for the intermediate in the mechanism.

• Step 1 – B is produced.

• Reverse of Step 1 – B is consumed.

• Step 2 – B is consumed.

BkBkAkdtBd

211

The SSA (Cont’d)

The SSA applied to the intermediate B.

21

1

211

kkAk

B

BkBkAk

0dtBd

SSA – The Final Step

Substitute the expression for the concentration of B into the rate law vp.

A

kkkk

dtPd

BkdtPd

21

12

2

Competing Reactions

Imagine a reaction with a competing side reaction.

v k A

v k A

b a

c b

A B

A C

The Reaction “Yield”

We can calculate the amount of material produced from the competing reactions.

J

nn

kk

kJ = the rate constant for the reaction J.

Activated Complex Theory

Consider the following bimolecular reaction

k

A B C Presume that the reaction proceeds through the

transition state?1

1

k k

k

A B AB P

The Activated Complex

The species temporarily formed by the reactant molecules – the activated complex.

A small fraction of molecules usually have the required kinetic energy to get to the transition state • The concentration of the activated complex is

extremely small.

Transition State Theory (TST)

TST pictures the bi-molecular reaction proceeding through the activated complex in a rapid-pre-equilibrium.

1

1

k k

k

A B AB P

d Pk AB

dt

TST (II)

From the thermodynamic equilibrium constant for the formation of AB↕

;

A B

AB

A B

A B

AB

pK

p p

p RT A p RT B

p RT AB

RTAB K

P

TST (IV)

Assume that we can obtain the Gibbs energy of activation from K↕.

2

lnG

B RTo

G RT K

k T RTk e

h P

The Eyring Equation!

TST (V)

Hence, substituting for the Gibbs energy of activation in terms of the enthalpy and entropy of activation.

2

SHB RT R

o

G H T S

k T RTk e e

h P

TST (VI)

Expressing the enthalpy of activation in terms of the activation energy.

2 bimolculargasphasereaction

unimoleculargasphase

or solutionreaction

a

a

H E RT

H E RT

TST (VII)

The final result for a bimolecular gas phase reaction.

22

aE SB RT R

o

k T RTk e e e

h P

2

2

2

a aE ESBRT R RT

o

SB R

o

k T RTk Ae e e

h Pk T RT

A eh P

TST (VIII)

The final result for a unimolecular gas phase reaction (or any solution reaction). aE S

B RT Rr o

k T RTk e e e

h P

1

1

a aE ESBRT R RT

r o

SB R

o

k T RTk Ae e e

h Pk T RT

A eh P